Limit of $\sqrt x$ as $x$ approaches $0$ What is the limit of $\sqrt x$ as $x$ approaches $0$? I asked a few people and they all gave me two different answers. Some said that the limit is $0$, and other say that the limit is not defined because the right-hand limit is $0$ while the left-hand limit is undefined. I'm confused. please help!!
 A: It may depend on definitions: for some, to even talk about
$$\;\lim_{x\to 0}f(x)\;$$     
the function must be defined on a complete neighborhood of $\;x_0\;$ , i.e. in an open interval of the form $\;(x_0-\epsilon\,,\,x_0+\epsilon)\;,\;\;\epsilon >0\;$ .
For others, which is what Daniel seems to point at, it is enough to talk of the function within its definition domain.
Thus, under the first point of view, we can not talk about
$$\lim_{x\to 0}\sqrt x\;\;\text{since this function's defined only for $\;x\ge 0\;$}$$
meaning: there is no "left" neighborhood of $\;0\;$ in the definition domain and thus we can only talk of the right-sided limit. It depends on the context.
A: From what I understand, the limit as $x$ approaches $0$ of $\sqrt{x}$ is in fact $0$. I think I understand where your confusion comes from. If I understand correctly then you will have been taught that
$$\lim_{x\rightarrow 0^-} \sqrt{x} = \lim_{x\rightarrow 0^+} \sqrt{x}$$
then
$$\lim_{x\rightarrow 0} \sqrt{x} $$ will exist or in other words, both the left and right side limit must exist for you to find both limits. However, the subtle "thing" one must understand is that the function has to be defined in the negative spectrum for $x$, i.e. it must exist in the negative domain to begin with. The square root function isn't defined in the negative domain and therefore, by definition, you cannot take the limit in that domain. This doesn't mean that the limit does not exist there, it just means that you cannot take a limit there to begin with. Thus, the limit at $0$ of the $f(x) = \sqrt{x}$ is in fact $0$ i.e.
$$\lim_{x\rightarrow 0} \sqrt{x} = 0$$
A: Many introductory calculus books define $\lim_{x\to a}f(x)$ under a premise that $f$ must be defined on an open interval containing $a$.  Since any open interval containing $0$ contains negative numbers, it would seem that $\lim_{x\to 0}\sqrt{x}$ is undefined.
However, the definition of a limit in the introductory calculus books is a narrow version of a definition normally used in mathematical analysis.  For example, according to Walter Rudin's "Principles of Mathematical Analysis" the limit can be defined at any limit point of the domain of $f$.  Since the domain of $f(x)=\sqrt{x}$ is the interval $[0,\infty)$, and $0$ is a limit point for this interval, then by such definition, $\lim_{x\to 0}\sqrt{x}$ is defined and is equal to $0$.
Of course, the concept of a "limit point" would require a discussion that could make an introductory calculus books not so introductory.  Hence the narrow definition of a limit.  My personal impression is that the introductory calculus books could easily be improved to include a definition of a limit that leaves no doubt about the fact that $\lim_{x\to 0}\sqrt{x}=0$. 
A: I think we should just understand that the requirement that the function be defined in an open neighborhood only makes sense if the point of interest (the point around which we seek to find the limit) is not an boundary point. by this it would almost suffice to say that if the point is a boundary point, then is the limit. Furthermore, we must also understand that if a function is defined at a certain point, then the limit as x approaches that point is simply the value of the function at that point. We must also acknowledge the fact that domains are determined by co-domains and so all the arguments made thus far, are premised on the fact that the co-domain of this function is the set of real numbers.  
